Duality Theory I: Basic Theory
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DT I c Gabriel Nagy Duality Theory I: Basic Theory Notes from the Functional Analysis Course (Fall 07 - Spring 08) This section contains an important conceptual discussion on duality, which in a special case encodes the interplay between a locally convex space and its (topological) dual. A. Dual Pairings Convention. Throughout this note K will be one of the fields R or C, and all vector spaces are over K. Definitions. Suppose one has two vector spaces X and Y.A dual pairing of X with Y is a bilinear1 map Φ : X × Y → K, which is non-degenerate, in the following sense: • If x ∈ X is such that Φ(x, y) = 0, ∀ y ∈ Y, then x = 0. • If y ∈ Y is such that Φ(x, y) = 0, ∀ x ∈ X , then y = 0. Equivalently, Φ can be represented by two2 linear injective maps #: X 3 x 7−→ x# ∈ Y0, (1) #: Y 3 y 7−→ y# ∈ X 0, (2) satisfying the equality x#(y) = y#(x), ∀ x ∈ X , y ∈ Y. (3) The bilinear map Φ is simply given as Φ(x, y) = x#(y) = y#(x). Example 1. Given an arbitrary vector space X , and a linear subspace F ⊂ X 0, which separates the points of X (i.e. if x1 6= x2, there exists f ∈ F with f(x1) 6= f(x2)), then the map Φ : X × F 3 (x, f) 7−→ f(x) ∈ K is a dual pairing. Example 2. We can specialize the above example to the following situation. Assume X is a locally convex topological vector space. In particular the topology T, which X is equipped with, is Hausdorff, and the dual space X ∗ = (X , T)∗ separates the points in X . Then the map Φ : X × X ∗ 3 (x, φ) 7−→ φ(x) ∈ K is a dual pairing. B. The Weak Dual Topologies Definitions. Assume now a dual pairing is given, defined by the maps (1) and (2) satis- fying (3). If we endow Y0 with the weak* topology, we can apply the pull-back construction 1 This means that Φ is linear in each variable. 2 Although the notation is abusive, it is very convenient! 1 (see LCVS II) using the map (1) to obtain a topology, denoted by w# on X . Specifically, this is the weakest topology which makes the map #: X → (Y0, w∗) continuous. Since the weak* topology on Y0 is Hausdorff locally convex, and the map # is linear and injective, the topology w# on X is Hausdorff locally convex. Similarly, one also constructs a Hausdorff locally convex topology w# on Y, as the weakest topology that makes the map #: Y → (X 0, w∗) continuous. The two topologies constructed above are called the weak topologies associated with the dual pairing. Remarks 1-2. Use the notations as above. 1. Convergence in (X , w#) and in (Y, w#) can be characterized as follows: w# (i) xλ −−→ x (in X ), if and only if one of the following equivalent conditions holds: # # xλ (y) → x (y), ∀ y ∈ Y, # # y (xλ) → y (x), ∀ y ∈ Y. w# (ii) yλ −−→ y (in X ), if and only if one of the following equivalent conditions holds: # # yλ (x) → y (x), ∀ x ∈ X , # # x (yλ) → x (y), ∀ x ∈ X . 2. In terms of seminorms: # X (i) the topology w on X is defined by the family PY = {py}y∈Y , given by # # py(x) = |y (x)| = |x (y)|, x ∈ X , y ∈ Y; # Y (ii) the topology w on Y is defined by the family PX = {px}x∈X , given by # # px(y) = |x (y)| = |y (x)|, x ∈ X , y ∈ Y; Theorem 1. Use the notations above. (i) For any x ∈ X , the linear functional x# : Y → K is continuous in the w# topology. (i’) If we equip the space Y∗ = (Y, w#)∗ with the weak* topology, the map #:(X , w#) 3 x 7−→ x# ∈ (Y∗, w∗) (4) is a topological linear isomorphism. (ii) For any y ∈ Y, the linear functional y# : X → K is continuous in the w# topology. (ii’) If we equip the space X ∗ = (X , w#)∗ with the weak* topology, the map #:(Y, w#) 3 y 7−→ y# ∈ (X ∗, w∗) (5) is a topological linear isomorphism. 2 Proof. By symmetry, it suffices to prove only (i) and (i’). Property (i) is quite trivial, by Remark 1. To prove (ii) we first remark that (4) is injective, by definition. Next we show that (4) is surjective. Suppose φ : Y → K is linear and w#-continuous, and let us prove that there exists (a unique) x ∈ X , such that φ = x#. Using Remark 2, the discussion from LCVS IV (Remark 3), there exist x1, . , xn ∈ X and t1, . , tn > 0, such that # # |φ(y)| ≤ t1|x1 (y)| + ··· + tn|xn (y)|, ∀ y ∈ Y. # Using Exercise ?? from HB, (applied to the seminorms pj(y) = tj|xj (y)|, j = 1, . , n), there exist linear functionals φ1, . , φn : Y → K, such that φ = φ1 + ··· + φn, and # |φj(y)| ≤ tj|xj (y)|, ∀ y ∈ Y, j = 1, . , n. # Fix for the moment j, and consider the linear functional ψj = tjxj : Y → K. Since |φj(y)| ≤ |ψj(y)|, ∀ y ∈ Y, we must have the inclusion Ker ψj ⊂ Ker φj, so X /Ker φj is a quotient of X /Ker ψj. Since ψj is linear, we must have 1 ≥ dim(Ker ψj) ≥ dim(Ker φj) ≥ 0, so either (i) dim(Ker ψj) = dim(Ker φj)(≤ 1), in which case Ker φj = Ker ψj, or (i) 1 = dim(Ker ψj) > dim(Ker φj) = 0, in which case Ker φj = X . In either case we get φj = αjψj, for some αj ∈ K (in case (ii), αj = 0). So now we have # φj = αjtjxj , j = 1, . , n, so # # # φ(y) = α1t1x1 (y) + ··· + α1t1x1 (y) = (α1t1x1 + ··· + αntnxn) (y), ∀ y ∈ Y, # i.e. φ = (α1t1x1 + ··· + αntnxn) . Having proven that (4) is a linear isomorphism, its continuity, as well as the continuity of ∗ # w # ∗ its inverse, are tautologic by Remark 1, since the condition xλ −→ x (in Y ) is equivalent to # # xλ (y) → x (y), ∀ y ∈ Y, w# which in turn is equivalent to xλ −−→ x (in X ). We now re-visit the two examples from sub-section A. Example 1. Given an arbitrary vector space X , and a linear subspace F ⊂ X 0, which separates the points of X , then, as pointed out in sub-section A, we have is a dual pairing between X and X . The weak topology w# on X is referred to as the weak F-topology, and is denoted by wF . The other weak topology w# on F is simply the restriction of the weak* topology (from X 0) to F. Example 2. If (X , T) is a locally convex topological vector space (thus T is Hausdorff), then, as discussed in Example 2 from sub-section A, we have a dual pairing between X and 3 ∗ # X . The topology w on X is simply the weak topology wT, implemented by T, which was introduced in LCVS IV. In particular (see LCVS IV), for every set S ⊂ X , one has the equalities T # conv S = conv SwT = conv Sw . (6) As before, the other topology w# on X ∗ is simply the weak* topology. As a consequence of Theorem 1, we also have a natural topological linear isomorphism # ∗ ∗ ∗ #:(X , wT) 3 x 7−→ x ∈ (X , w ) , defined by x#(φ) = φ(x), ∀ x ∈ X , φ ∈ X ∗. Exercise 1*. Equip the direct sum X = L with the locally convex sum topology N K T. Show that the weak topology wT is strictly weaker than T. Conclude that there does not exist locally convex spaces Y, such that (X , T) is topologically linearly isomorphic to (Y∗, w∗). Q Exercise 2. Let I be some non-empty set. Consider the spaces Xprod = K and L I Xsum = I K. 3 (i) Prove that the map Φ : Xprod × Xsum → K, defined by X Φ(x, y) = xiyi, ∀ x = (xi)i∈I ∈ Xprod, y = (yi)i∈I ∈ Xsum, i∈I establishes a dual pairing. # (ii) Prove that the topology w on Xprod coincides with the product topology Tprod. # (iii) Prove that the topology w on Xsum is strictly weaker than the locally convex sum ∗ topology. Therefore we only have an algebraic linear isomorphism Xsum ' (Xprod, Tprod) . C. Polars and the Bipolar Theorem As we have already seen in Example 2, the closure of convex hulls depends only on the interaction between the ambient space and its (topological) dual. Therefore, it is expected that the operation of taking closed convex hulls to admit an “abstract” characterization, within the framework of dual pairs. Definitions. Suppose a dual pairing between two vector spaces X and Y is given, defined by the maps (1) and (2) satisfying (3). For a non-empty set A ⊂ X , we define its polar (in Y) to be the set: A◦ = {y ∈ Y : Re y#(a) ≤ 1, ∀ a ∈ A}. Similarly, for a non-empty set B ⊂ Y, we define its polar (in X ) to be the set: B◦ = {x ∈ X : Re x#(b) ≤ 1, ∀ b ∈ B}. Exercise 3. Prove that, using the notations as above, for two non-empty sets A ⊂ B ⊂ X , one has the reverse inclusion A◦ ⊃ B◦. 3 The sum is clearly finite, since yi 6= 0 only for finitely many i ∈ I. 4 Proposition 2. With the notations as above, if A ⊂ X is non-empty, then A◦ is convex, w#-closed in Y, and contains 0. Likewise, if B ⊂ Y is non-empty, then B◦ is convex, w#-closed in X , and contains 0. Proof. By symmetry, we only need to prove the first statement. The fact that A◦ contains 0 is trivial.